
What mass of butter, which has a usable energy content of\[6.0Cal/g\left( { = 6000cal/g} \right)\] , would be equivalent to the change in gravitational potential energy of a \[73.0kg\] man who ascends from sea level to the top of Mt. Everest, at elevation\[8.84km\] ? Assume that the average \[g\] for the ascent is\[9.80m/{s^2}\].
Answer
583.8k+ views
Hint:- In this question, the work done by the man for climbing the top of Mt. Everest is equal to the energy given by the butter. We can find the work done by using the formula $work = mass \times acceleration \times height$.
Complete step-by-step solution:
According to the question, the mass of the man is \[{m_{man}} = 73kg\]. The height of the Mt. Everest is \[h = 8.84km\]. So, the work which the man has to do to climb to the top of the Mt. Everest is given by
\[W = {m_{man}}gh\]
(As $work = force \times displacement$ or we can write $work = mass \times acceleration \times height$)
Putting the values of\[{m_{man}} = 73kg\] , \[g = 9.8m/{s^2}\] and \[h = 8.84km = 8840m\] in the above equation, we get-
\[\begin{array}{*{20}{l}}
{ \Rightarrow W = 73 \times 9.80 \times 8840} \\
{ \Rightarrow W = 6.32 \times {{10}^6}J}
\end{array}\]
So, the man needs to work $6.32 \times {10^6}J$ to climb the top of Mt. Everest.
Now, this energy is equal to the energy given by butter.
Let the mass of the butter is ${m_{butter}}$ and the usable energy of the butter is ${U_{butter}}$ , then-
$W = {m_{butter}}{U_{butter}}$
\[6.32 \times {10^6} = {m_{butter}}25200\]
(whereas $1cal = 4.2J$so, ${U_{butter}} = 6000 \times 4.2 = 25200J$ )
Now, simplifying the equation, we get-
\[
{m_{butter}} = \dfrac{{6.32 \times {{10}^6}}}{{25200}} \\
{m_{butter}} = 250.79g \\
\]
Thus, the amount of butter needed is $250.79g$.
Therefore, $250.79g$ of butter, which has a usable energy content of\[6.0Cal/g\left( { = 6000cal/g} \right)\] , would be equivalent to the change in gravitational potential energy of a \[73.0kg\] man who ascends from sea level to the top of Mt. Everest, at elevation\[8.84km\].
Note:- In this question, we have to keep in mind that for calculating the work done, all the quantities should be in the MKS system (as height of the Mt. Everest is in Kilometres). Similarly, the usable energy of butter is given in calories/grams. But the work done is in Joules. So, we have to change the usable energy of the butter into joules/grams by using the relation of calorie and joule i.e. $1cal = 4.2J$.
Complete step-by-step solution:
According to the question, the mass of the man is \[{m_{man}} = 73kg\]. The height of the Mt. Everest is \[h = 8.84km\]. So, the work which the man has to do to climb to the top of the Mt. Everest is given by
\[W = {m_{man}}gh\]
(As $work = force \times displacement$ or we can write $work = mass \times acceleration \times height$)
Putting the values of\[{m_{man}} = 73kg\] , \[g = 9.8m/{s^2}\] and \[h = 8.84km = 8840m\] in the above equation, we get-
\[\begin{array}{*{20}{l}}
{ \Rightarrow W = 73 \times 9.80 \times 8840} \\
{ \Rightarrow W = 6.32 \times {{10}^6}J}
\end{array}\]
So, the man needs to work $6.32 \times {10^6}J$ to climb the top of Mt. Everest.
Now, this energy is equal to the energy given by butter.
Let the mass of the butter is ${m_{butter}}$ and the usable energy of the butter is ${U_{butter}}$ , then-
$W = {m_{butter}}{U_{butter}}$
\[6.32 \times {10^6} = {m_{butter}}25200\]
(whereas $1cal = 4.2J$so, ${U_{butter}} = 6000 \times 4.2 = 25200J$ )
Now, simplifying the equation, we get-
\[
{m_{butter}} = \dfrac{{6.32 \times {{10}^6}}}{{25200}} \\
{m_{butter}} = 250.79g \\
\]
Thus, the amount of butter needed is $250.79g$.
Therefore, $250.79g$ of butter, which has a usable energy content of\[6.0Cal/g\left( { = 6000cal/g} \right)\] , would be equivalent to the change in gravitational potential energy of a \[73.0kg\] man who ascends from sea level to the top of Mt. Everest, at elevation\[8.84km\].
Note:- In this question, we have to keep in mind that for calculating the work done, all the quantities should be in the MKS system (as height of the Mt. Everest is in Kilometres). Similarly, the usable energy of butter is given in calories/grams. But the work done is in Joules. So, we have to change the usable energy of the butter into joules/grams by using the relation of calorie and joule i.e. $1cal = 4.2J$.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

